Confidence Intervals: Passing to logarithm So one of my problems says to take an answer from the previous part of the problem and "pass it to logarithm" to find a different one and interpret it.  Basically, we had to find a CI for $2^{\alpha_2}/2^{\alpha_1}$ and now we need to find one for $\alpha_2-\alpha_1$ by using logarithms.  My question then is, do I take the logarithm of the entire interval or just the pivot around which I'm finding my CI?  And then how would I interpret the new CI?
 A: You can obtain the new parameter of interest by taking your previous parameter to the logarithm base-2.  Since this logarithm is a strictly increasing function, you can obtain a confidence interval for the latter quantity by applying the logarithm to the boundaries of your previous confidence interval.  To see this, we note that for any random data vector $\mathbf{X}$ and bound functions $L$ and $U$ we have:
$$\mathbb{P} \Big( L(\mathbf{X}) \leqslant \frac{2^{\alpha_2}}{2^{\alpha_1}} \leqslant U(\mathbf{X}) \Big) 
= \mathbb{P}(\log_2 L(\mathbf{X}) \leqslant \alpha_2 - \alpha_1 \leqslant \log_2 U(\mathbf{X})).$$
Thus, your new confidence interval is:
$$\text{CI} = [\log_2 L(\mathbf{x}), \log_2 L(\mathbf{x})].$$
Note that while this gives you a confidence interval for the new parameter, it will not generally be an optimal confidence interval (i.e., the shortest interval with the required confidence level).  If you want to obtain an optimal confidence interval for the new parameter then you would have to go back to the underlying interval formula and optimise the lower-tail probability for the interval.
